Question

Como calcular a tangente de 75°?

Answer 1

tg75=(tag45+tg30)/(1-tg45.tg30) 
tan75=(tag30=(√3/3+1)/(1-√3/3 .1)= ((√3+3)/3)/(3-√3)/3)= 

((√3+3))/(3-√3)= racionalizando, multiplicando por (√3+3) 

fica 

((√3+3))²/((√3+3)(3-√3)=(12+6√3)/( 9-3)=(12+6√3)/6= (2+√3)/1 

tag 75=(12+6√3)/6 ou 

tag75=(2+√3)/1 

=3.732051

Answer 2

 Olá Arielalves!

 Note que: $75^o = 30^o + 45^o$.

 Se assim como eu, tiveres dificuldades de memorizar muitas fórmulas sugiro que memorize as básicas, pois é o que faço! Rs

 Sabemos que $\tan \, \alpha = \frac{\sin \, \alpha}{\cos \, \alpha}$; isto posto, temos que:

$\tan \, 75^o = \\\\ \frac{\sin \, 75^o}{\cos \, 75^o} = \\\\ \frac{\sin (30^o + 45^o)}{\cos (30^o + 45^o)} = \\\\ (...)$

 Faltou dizer que: $\begin{cases}\sin (a + b) = \sin \, a \cdot \cos \, b + \sin \, b \cdot \cos\, a \\ \cos (a + b) = \cos \, a \cdot \cos \, b - \sin \, a \cdot \sin \, b \end{cases}$

 Segue,

$\\ \frac{\sin (30^o + 45^o)}{\cos (30^o + 45^o)} = \\\\ \frac{\sin 30^o \cos 45^o + \sin 45^o \cdot \cos 30^o}{\cos 30^o \cdot \cos 45^o - \sin 30^o \cdot \sin 45^o} = \\\\ \frac{\frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2}} = \\\\ \\ \frac{\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}}{\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}} = \\\\ \frac{\sqrt{2} + \sqrt{6}}{4} \div \frac{\sqrt{6} - \sqrt{2}}{4} = \\\\ \frac{\sqrt{2} + \sqrt{6}}{4} \times \frac{4}{\sqrt{6} - \sqrt{2}} = \\\\ \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} =$

 Racionalizando,

$\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} \times \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}}= \\\\ \frac{6 + 2\sqrt{12} + 2}{6 - 2} = \\\\ \frac{8 + 4\sqrt{3}}{4} = \\\\ \boxed{2 + \sqrt{3}}$